ABSTRACT (original article): Whilst the complexity of acquiring knowledge of a quantum state has been extensively studied in the fields of quantum tomography and quantum learning, a physical understanding of the role of knowledge in quantum thermodynamics is lacking. Knowledge is central to thermodynamics as exemplified by the Maxwell’s demon thought experiment where a demonic agent is able to extract paradoxical amounts of work--reconciled by the cost of acquiring this knowledge. In this work, we address this gap by extending known unitary models for the acquisition of quantum knowledge to account for the resources an agent has access to; in particular the dimension of their probe system. We view an agent’s knowledge of a quantum state as their ability to reconstruct it unitarily. In our model an agent correlates an unknown d-dimensional system with copies of a lower k-dimensional probe, which they then use to reconstruct estimates of the state in d-dimensional memories. We find that this framework is a unitary representation of coarse-grained POVMs, connecting thermodynamics and state estimation. As applications it allows us to investigate the role of knowledge in work extraction and state transformation. CITATION (original article): Jake Xuereb, A. de Oliveira Junior, Fabien Clivaz, Pharnam Bakhshinezhad, Marcus Huber (2024), What resources do agents need to acquire knowledge in Quantum Thermodynamics? arXiv:2410.18167. https://doi.org/10.48550/arXiv.2410.18167
How hard is it to reconstruct the state of an unknown system given access to correlations with a probe? In this work the authors have examined this question, creating a bridge between state estimation and quantum thermodynamics. In the below code the authors use Mathematica to explore the exact reconstruction of the state of an unknown qutrit system from correlations form with 3 qubit probes. The qubit probes are used to construct three approximate representation of the state of the system in quantum memories. These approximate representations are referred to as estimate states. Given access to these three qutrit estimate states, does there exist a unitary that exactly reconstructs the unknown state? The below code shows that this is true when a marginal majorisation condition is satisfied. In the code the authors begin by finding the set of states for which the condition is satisfied and then continue to construct an example of a unitary which carries out the desired exact reconstruction. To do this, the authors take this 27 dimensional space and break it down into a direct sum of 9, 3 dimensional spaces over which they find a bistochastic matrix which carries out the desired transformation.
QI “Quantum Information” package: https://github.com/rogercolbeck/QI
QI Manual: https://www-users.york.ac.uk/~rc973/QI/QI_manual.pdf
QI Manual: https://www-users.york.ac.uk/~rc973/QI/QI_manual.pdf
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<<QI`
The Unknown Target State
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The Estimates one obtains by running our protocol with access to a qubit probe .
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ω
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Symmetrised Estimate State
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Symmetrised[p0_,p1_,p2_]=(1/2)*[p0,p1,p2]+IdentityMatrix[3]3;
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Function to convert Lexicographic ordering to computational basis
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LexToNum[i_,j_,k_]=9*i+3*k+j+1;(*BasisKetGeneratorandHammingWeightDistantGHZLikeSubspacesGenerator*)BasisKet[i_]:=IdentityMatrix[27][[i]]Subspace[k_,l_]:=Table[BasisKet[LexToNum[j,Mod[k+j,3],Mod[l+j,3]]],{j,0,2}];
Product of Estimate States
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(*Switcheroo*)
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OmegaBar[p0_,p1_,p2_]=KroneckerProduct[[p0,p1,p2],[p0,p1,p2],[p0,p1,p2]];(*Sortthevectorofeigenvaluesforagivenchoiceofvalues*)Sorted[p0_,p1_,p2_]:=Sort[Diagonal[OmegaBar[p0,p1,p2]],Greater];OmegaBarDesc[p0_,p1_,p2_]:=DiagonalMatrix[Sorted[p0,p1,p2]];
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Construct Marginal For a given Choice of Values with Global Sorting
MarginalDesc[p0_,p1_,p2_]:=PartialTrace[OmegaBarDesc[p0,p1,p2],3,9,2]MarginalDescSums[p0_,p1_,p2_]:={Diagonal[MarginalDesc[p0,p1,p2]][[1]],Diagonal[MarginalDesc[p0,p1,p2]][[1]]+Diagonal[MarginalDesc[p0,p1,p2]][[2]],Total[Diagonal[MarginalDesc[p0,p1,p2]]]}
Let’ s check when a given qutrit target state will have estimate states which allow for reconstruction i . e . satisfy the majorisation condition
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Ineq[p0_,p1_,p2_]:={MarginalDescSums[p0,p1,p2][[1]]>=p0,MarginalDescSums[p0,p1,p2][[2]]>=p0+p1};
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True[Ineq[3/6,2/6,1/6]]
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True[{True,True}]
Let’ s check for an equally spaced hamiltonian
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MarginalKet[i_]:=IdentityMatrix[3][[i]]
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MarginalProj[i_]:=Outer[Times,MarginalKet[i],MarginalKet[i]]
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H[E0_,E1_,E2_]=E0*MarginalProj[1]+E1*MarginalProj[2]+E2*MarginalProj[3]
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ThermalPops[β_,E0_,E1_,E2_]=Diagonal[MatrixExp[-β*H[E0,E1,E2]]/Tr[MatrixExp[-β*H[E0,E1,E2]]]];
Globally Sorted Inequality 1
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Plot[{Diagonal[MarginalDesc[ThermalPops[β,0,1,2][[1]],ThermalPops[β,0,1,2][[2]],ThermalPops[β,0,1,2][[3]]]][[1]],ThermalPops[β,0,1,2][[1]]},{β,0,10},PlotStyle->{{Red,Dashed,Thick},{Green}},PlotRange->All]
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Global Sorting Inequality 2
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Plot[{Diagonal[MarginalDesc[ThermalPops[β,0,1,2][[1]],ThermalPops[β,0,1,2][[2]],ThermalPops[β,0,1,2][[3]]]][[1]]+Diagonal[MarginalDesc[ThermalPops[β,0,1,2][[1]],ThermalPops[β,0,1,2][[2]],ThermalPops[β,0,1,2][[3]]]][[2]],ThermalPops[β,0,1,2][[1]]+ThermalPops[β,0,1,2][[2]]},{β,0,5},PlotStyle->{Red,Green},PlotRange->All]
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Very cool there should exist a Bistochastic Matrix which when applied to each subspace gives us this transformation - Let’ s now construct the subspaces and find this Bistochastic Matrix
Vectors of Populations in Each of the Subspaces
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RVec[k_,l_,p0_,p1_,p2_]:=Table.OmegaBarDesc[p0,p1,p2].Subspace[k,l][[j]],{j,1,3}
Subspace[k,l][[j]]
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AllRVecs[p0_,p1_,p2_]:={RVec[0,0,p0,p1,p2],RVec[0,1,p0,p1,p2],RVec[0,2,p0,p1,p2],RVec[1,0,p0,p1,p2],RVec[2,0,p0,p1,p2],RVec[1,1,p0,p1,p2],RVec[1,2,p0,p1,p2],RVec[2,1,p0,p1,p2],RVec[2,2,p0,p1,p2]};
The 6 3 X3 Permutation Matrices
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PM[l_]:=PermutationMatrix[Permutations[{1,2,3}][[l]]]
A Bistochastic Matrix is a convex mixture of permutation matrices
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BiStoch[a0_,a1_,a2_,a3_,a4_,a5_]=FullSimplify[a0*PM[1]+a1*PM[2]+a2*PM[3]+a3*PM[4]+a4*PM[5]+a5*PM[6],{a0+a1+a2+a3+a4+a5+a6==1,a0>=0,a1>=0,a2>=0,a3>=0,a4>=0,a5>=0}]
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{{a0+a1,a2+a3,a4+a5},{a2+a4,a0+a5,a1+a3},{a3+a5,a1+a4,a0+a2}}
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{{a0+a1,a2+a3,a4+a5},{a2+a4,a0+a5,a1+a3},{a3+a5,a1+a4,a0+a2}}//MatrixForm
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a0+a1 | a2+a3 | a4+a5 |
a2+a4 | a0+a5 | a1+a3 |
a3+a5 | a1+a4 | a0+a2 |
Bistochastic Matrix Applied to the sum of the population vectors
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FirstEntryMarginal[a0_,a1_,a2_,a3_,a4_,a5_,p0_,p1_,p2_]:=Total[Flatten[Table[(BiStoch[a0,a1,a2,a3,a4,a5].RVec[k,l,p0,p1,p2])[[1]],{k,0,2},{l,0,2}]]];SecondEntryMarginal[a0_,a1_,a2_,a3_,a4_,a5_,p0_,p1_,p2_]:=Total[Flatten[Table[(BiStoch[a0,a1,a2,a3,a4,a5].RVec[k,l,p0,p1,p2])[[2]],{k,0,2},{l,0,2}]]];ThirdEntryMarginal[a0_,a1_,a2_,a3_,a4_,a5_,p0_,p1_,p2_]:=Total[Flatten[Table[(BiStoch[a0,a1,a2,a3,a4,a5].RVec[k,l,p0,p1,p2])[[3]],{k,0,2},{l,0,2}]]];
Equations for a given set of values fixed with the sorting fixed
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EquationsNum[p0_,p1_,p2_]:=FirstEntryMarginal[a0,a1,a2,a3,a4,a5,p0,p1,p2]==p0&&SecondEntryMarginal[a0,a1,a2,a3,a4,a5,p0,p1,p2]==p1&&ThirdEntryMarginal[a0,a1,a2,a3,a4,a5,p0,p1,p2]==p2&&a0+a1+a2+a3+a4+a5==1&&a0>=0&&a1>=0&&a2>=0&&a3>=0&&a4>=0&&a5>=0;
Let’ s see if there’ s a Bistchochastic Matrix for the example ={3/6,2/6,1/6} we chose earlier
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FindInstance[EquationsNum[3/6,2/6,1/6],{a0,a1,a2,a3,a4,a5},Reals,3]
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a0,a10,a2,a30,a40,a50
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B1=BiStoch,0,,0,0,0
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,,0,,,0,{0,0,1}
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,,0,,,0,{0,0,1}//MatrixForm
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9 10 | 1 10 | 0 |
1 10 | 9 10 | 0 |
0 | 0 | 1 |
CITE THIS NOTEBOOK
CITE THIS NOTEBOOK
What resources do agents need to acquire knowledge in quantum thermodynamics?
by Jake Xuereb, A. de Oliveira Junior, Fabien Clivaz, Pharnam Bakhshinezhad, Marcus Huber
Wolfram Community, STAFF PICKS, November 5, 2024
https://community.wolfram.com/groups/-/m/t/3314686
by Jake Xuereb, A. de Oliveira Junior, Fabien Clivaz, Pharnam Bakhshinezhad, Marcus Huber
Wolfram Community, STAFF PICKS, November 5, 2024
https://community.wolfram.com/groups/-/m/t/3314686